Answer:
a) 43 m b) 55 m
Step-by-step explanation:
a) From question at t = 0, initial velocity [tex]V_{o}[/tex] = 15 m/s
Using equation of motion, [tex]S = V_{o}t + \frac{1}{2} at^{2}[/tex] ; at t = 2 secs , a = 4 m/[tex]s^{2}[/tex]
S = (15 x 2) + (0.5 X 4 x [tex]2^{2}[/tex])
S = 30 + 8 = 38 m , Therefroe;
car is (38 + 5)m from the sign post
car is 43 m from the sign post at t = 2 secs
b) Also from equation of motion, [tex]V^{2} = V_{o} ^{2} + 2aS[/tex]
[tex]25^{2} = 15^{2}[/tex] + (2 x 4 x S)
625 - 225 = 8S
S = 50 m
Car is (50 + 5) m from the sign post
Car is 55 m from the sign post at V = 25 m/s
